Let $a,$ $b,$ and $c$ be positive real numbers such that $a > b$ and $a + b + c = 4.$

Find the minimum value of $4a + 3b + \frac{c^3}{(a - b)b}.$

How do I find the minimum value of this? Any help I will appreciate, and I will get back to you extremely quickly. Thank you!

HighSchoolDx May 12, 2021

#1**+4 **

AM-GM can likely help you here lol

the equation would go $a + \dfrac{c^3}{(a - b)b} \geq 3 \sqrt[3]{ \dfrac{(a - b)bc^3}{(a - b)b}} = 3c.$

this'd imply that, $4a + 3b + \frac{c^3}{(a - b)b} \geq 3a + 3b + 3c$

the rhs is 3(a + b + c), or 12 - thus, **12 **is the local minimum.

CentsLord May 12, 2021

#2**+2 **

THANK you!!!

apologies for the long time for reply... i was on vacation. I will be sure to study up on my artihmetic-geometric mean inequalities ðŸ˜…

HighSchoolDx
May 16, 2021